Alongside the announcement of Apple News Plus, Apple‘s keynote today also released information on Apple Arcade, the company’s newly developed platform for gaming.
Much to the success of its App store and the 300,000+ of free and paid games found on it, Apple’s new Arcade platform will allow subscriptions to a gaming platform that include over 100 new and exclusive video games, of which are hand-picked for its subscribers. Details of the service include how Apple is contributing to the development costs with the creators — some of which include developers like Hironobu Sakaguchi (Final Fantasy), Ken Wong (Monument Valley), Will Wright (The Sims) and more.
A highlight feature of Apple Arcade is that the platform is cross-device, so players can start the game on their iPad, continue on their iPhone while commuting to work, and finishing the chapter or continue beyond their save point once they get home on their Mac or Apple TV. Lastly, privacy importance is spotlighted once again, with the company stating Arcade will not have ads or in-app purchases. Lastly, gaming will be made available offline as well for the ultimate access and ease of play for the catalog of titles.
Phil Schiller summarizes the service’s benefits:
“The App Store is the world’s biggest and most successful game platform. Now we are going to take games even further with Apple Arcade, the first game subscription service for mobile, desktop and the living room. We are working with some of the most innovative game developers in the world to create over 100 new and exclusive games to play across iPhone, iPad, Mac and Apple TV. Apple Arcade games will be great for families, respect user privacy and will not have ads or require any additional purchases. We think players of all ages are going to love Apple Arcade."
Apple Arcade is launching in Fall 2019 with 150+ countries coming on board as a new tab in the App Store. No pricing is revealed for now, but learn more over at Apple.com.
Recently, Xbox Live is coming to iOS as well as Nintendo Switch and Android.